Modern Math

Set Theory

Medium

If among $200$ students, $105$ like pizza and $134$ like burger, then the number of students who like only burger can possibly be

$93$

$26$

$96$

$23$

✅ Correct Option: 1

We need to organize the information about students who like pizza and burger using a Venn diagram approach.

Given information:

Total students = 200

Students who like pizza = 105

Students who like burger = 134

The key insight is that some students like both pizza and burger, so we can't just add 105 + 134 = 239 (that would be more than 200 students!).

Let us define our variables clearly:

Let $m$ = number of students who like both pizza and burger

Let $n$ = number of students who like neither pizza nor burger

Now we can break down all students into 4 distinct groups:

Only pizza = $105 - m$ (total pizza lovers minus those who like both)

Both pizza and burger = $m$

Only burger = $134 - m$ (total burger lovers minus those who like both)

Neither = $n$

Since these 4 groups must add up to the total number of students:

(Only pizza) + (Both) + (Only burger) + (Neither) = 200

Substituting our expressions:

$(105 - m) + m + (134 - m) + n = 200$

Simplifying:

$105 - m + m + 134 - m + n = 200$

$239 - m + n = 200$

$m - n = 39$

For this to make sense, we need:

$m \geq 0$ (can't have negative students)

$n \geq 0$ (can't have negative students)

$m \leq 105$ (can't have more students liking both than total pizza lovers)

From $m - n = 39$ and $n \geq 0$ :

$m \geq 39$

From $m \leq 105$ and $m \geq 39$ :

$39 \leq m \leq 105$

Students who like only burger = $134 - m$

When $m = 39$ : Only burger = $134 - 39 = 95$

When $m = 105$ : Only burger = $134 - 105 = 29$

Therefore, the number of students who like only burger lies in the range $[29, 95]$ .

Any value between 29 and 95 (inclusive) is a possible answer for students who like only burger.

Key Learning: In overlapping sets problems, always account for the intersection (students who like both) to avoid double-counting!

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If among 200200200 students, 105105105 like pizza and 134134134 like burger, then the number of students who like only burger can possibly be

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If among $200$ students, $105$ like pizza and $134$ like burger, then the number of students who like only burger can possibly be